hdu 5878 丑数 STL（pair，priority_queue）

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原来把算的数存起来需要时再输出就叫打表，先写个代码找找maxn

http://blog.csdn.net/qq_22497299/article/details/52565561

51nod1010：https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1010

一段只过部分数据的代码，可能常数有点大：

#include<iostream>#include<algorithm>#include<set>#include<queue>using namespace std;typedef long long ll;const int coeff[3]={2,3,5};set<ll> s;const int MAXN = 1e4 + 1e3;int main(){    priority_queue<ll, vector<ll>, greater<ll> > pq;    pq.push(1);    s.insert(1);    for(int i=0;i<MAXN;i++){        ll x=pq.top();        pq.pop();        for(int j=0;j<3;j++){            ll x2=x*coeff[j];            if(!s.count(x2)){                s.insert(x2);pq.push(x2);            }        }    }    int T;    cin>>T;    while(T--){        ll a;        cin>>a;        if(a==1){            cout<<"2"<<endl;            continue;        }        set<ll>::iterator it=s.find(a);        while(s.find(a)==s.end()){            a++;            it=s.find(a);        }        cout<<*it<<endl;    }    return 0;}

实际上最好二分：摘自http://blog.csdn.net/f_zyj/article/details/52077222

#include <iostream>#include <cstdio>#include <queue>using namespace std;typedef unsigned long long ull;const int MAXN = 1e4 + 1e3;/* * Ugly Numbers * Ugly numbers are numbers whose only prime factors are 2, 3 or 5. * 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... */typedef pair<ull, int> node_type;ull result[MAXN];void init(){    priority_queue<node_type, vector<node_type>, greater<node_type>> Q;    Q.push(make_pair(1, 2));    for (int i = 0; i < MAXN; i++)    {        node_type node = Q.top();        Q.pop();        switch (node.second)        {            case 2:                Q.push(make_pair(node.first * 2, 2));            case 3:                Q.push(make_pair(node.first * 3, 3));            case 5:                Q.push(make_pair(node.first * 5, 5));        }        result[i] = node.first;    }    return ;}/* *  传入参数必须l <= h *  假设a数组已经按从小到大排序 *  返回值l总是合理的 */int bs(ull a[], int l, int h, ull v){    int m;    while (l < h)    {        m = (l + h) >> 1;        if (a[m] < v)        {            l = m + 1;        }        else        {            h = m;        }    }    return l;}int main(int argc, const char * argv[]){    freopen("input.txt", "r", stdin);//    freopen("input.txt", "w", stdin);    init();    int T;    cin >> T;    ull n;    while (T--)    {        cin >> n;        int key = bs(result, 1, MAXN - 1, n);        cout << result[key] << '\n';    }    return 0;}

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